What is inverse oflaplace 1/((s+1)(s+3)^2) ?

The answer to inverse oflaplace 1/((s+1)(s+3)^2) is 1/4 e^{-t}-1/4 e^{-3t}-(e^{-3t}t)/2

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inverse oflaplace 1/((s+1)(s+3)^2)

inverse laplace

\frac{1}{( s + 1 ) ( s + 3 ) ^{2}}

\frac{1}{4} e^{- t} - \frac{1}{4} e^{- 3 t} - \frac{e ^{- 3 t} t}{2}

Solution steps

L^{- 1} {\frac{1}{( s + 1 ) ( s + 3 ) ^{2}}}

Take the partial fraction of

\frac{1}{( s + 1 ) ( s + 3 ) ^{2}} : \frac{1}{4 ( s + 1 )} - \frac{1}{4 ( s + 3 )} - \frac{1}{2 ( s + 3 ) ^{2}}

= L^{- 1} {\frac{1}{4 ( s + 1 )} - \frac{1}{4 ( s + 3 )} - \frac{1}{2 ( s + 3 ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{4 ( s + 1 )}} - L^{- 1} {\frac{1}{4 ( s + 3 )}} - L^{- 1} {\frac{1}{2 ( s + 3 ) ^{2}}}

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

L^{- 1} {\frac{1}{4 ( s + 3 )}} : \frac{1}{4} e^{- 3 t}

L^{- 1} {\frac{1}{2 ( s + 3 ) ^{2}}} : \frac{e ^{- 3 t} t}{2}

= \frac{1}{4} e^{- t} - \frac{1}{4} e^{- 3 t} - \frac{e ^{- 3 t} t}{2}

\int \frac{x ^{3} + 6 x ^{2} + 2 x + 3}{x ^{2} + 6 x + 1} d x

\int \frac{1}{49 e ^{- 7 x} + e ^{7 x}} d x

\frac{d y}{d x} = e^{x^{2}}

(ln (x))^{4}

What is inverse oflaplace 1/((s+1)(s+3)^2) ?
The answer to inverse oflaplace 1/((s+1)(s+3)^2) is 1/4 e^{-t}-1/4 e^{-3t}-(e^{-3t}t)/2